enter image description here

The answer is in terms of k. I tried finding a pattern between the consecutive terms of the series but could find none. Also I feel like there might be a systematic way for solving this that I do not know. My question is how would you solve this problem?

Thank you!


I presume that the Wikepedia site Ross Milikan links to says this: If we try a solution of the form $a_k= a^k$, for some constant k, then $a_{k-1}= a^{k-1}= \frac{a^k}{a}$ and $a_{k-2}= a^{k-2}= \frac{a^k}{a^2}$ so that $a_k- 4a_{k-1}+ 3a_{k-2}= a^k- \frac{4a^k}{a}+ \frac{4a^k}{k^2}= 0$. Dividing by $a^k$ (which, for a non-zero, is never 0) we have $1- \frac{4}{a}+ \frac{3}{a^2}= 0$. Multiplying by $a^2$ this becomes $a^2- 4a+ 3= (a- 3)(a- 1)= 0$ which has roots for a= 3 and a= 1. So both $a_k= 3^k$ and $a_k= 1^k= 1$ are solutions. And it is then easy so show that $a_k= C_13^k+ C_2$, where $C_1$ and $C_2$ can be any constants, is the "general" solution.

  • $\begingroup$ I believe you mean "for some constant $a$". $\endgroup$ – YoTengoUnLCD Jul 15 '16 at 21:40

Let $A(x)=\sum\limits_{k=1}^{\infty}a_kx^{k}$ be the generating function of the sequense $\{a_k\}$.

Now, by summing up from $k=3$ to $n$ the recursive relation we have that

$$ \sum\limits_{k=3}^{n}a_kx^k-4\sum\limits_{k=3}^{n}a_{k-1}x^k+3\sum\limits_{k=3}^{n}a_{k-2}x^k=0\Longrightarrow \sum\limits_{k=3}^{n}a_kx^k-4x\sum\limits_{k=2}^{n-1}a_{k}x^k+3x^2\sum\limits_{k=1}^{n-2}a_{k}=0 $$

Let $n$ go to $\infty$

$$ \sum\limits_{k=3}^{\infty}a_kx^k-4x\sum\limits_{k=2}^{\infty}a_{k}x^k+3x^2\sum\limits_{k=1}^{\infty}a_{k}=0\Longrightarrow [A(x)-a_2x^2-a_1x]-4x[A(x)-a_1x]+3x^2A(x)=0\Longrightarrow [A(x)-2x^2-x]-4x[A(x)-x]+3x^2A(x)=0\Longrightarrow (3x^2-4x+1)A(x)=2x^2+2x\Longrightarrow A(x)=\cfrac{2x^2+2x}{3x^2-4x+1} $$

and so by partially decomposition the fraction $\cfrac{2x^2+2x}{3x^2-4x+1}$ we have that


The generating functions of $\cfrac{1}{1-3x}$ and $\cfrac{1}{1-x}$ is $\sum\limits_{k=1}^{\infty}3^kx^k$ and $\sum\limits_{k=1}^{\infty}x^k$

Hence that,

$$A(x)=\sum\limits_{k=1}^{\infty}3^kx^k + \sum\limits_{k=1}^{\infty}x^k=\sum\limits_{k=1}^{\infty}(3^k+1)x^k$$

So, $ \ \ \ a_k=3^k+1$, $\ \ k\in\Bbb N$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.